Ask a Doubt
Request a call back

Join NOW to get access to exclusive study material for best results

JEE Class main Answered

Please explain the following
question image
Asked by Balbir | 24 Dec, 2018, 08:09: AM
answered-by-expert Expert Answer
Let the solubility of Ca(OH)2 be 'x'  mol/lit
left square bracket Ca to the power of 2 plus end exponent right square bracket space equals space straight x space mol divided by lit left square bracket OH to the power of minus right square bracket space equals space 2 straight x space mol divided by lit straight K subscript sp space space equals space left square bracket Ca to the power of 2 plus end exponent right square bracket space left square bracket OH to the power of minus right square bracket space space space space space space space space space equals space 4.42 cross times 10 to the power of negative 5 end exponent space straight x open parentheses 2 straight x close parentheses squared space equals space 4.42 cross times 10 to the power of negative 5 end exponent space straight x cubed space equals open parentheses fraction numerator space 4.42 cross times 10 to the power of negative 5 end exponent space over denominator 4 end fraction close parentheses straight x space space equals space open parentheses fraction numerator space 4.42 cross times 10 to the power of negative 5 end exponent space over denominator 4 end fraction close parentheses to the power of bevelled 1 third end exponent straight x equals space 2.23 cross times 10 to the power of negative 2 end exponent space mol divided by lit Now comma Ca left parenthesis OH right parenthesis subscript 2 space equals space left square bracket Ca to the power of 2 plus end exponent right square bracket space equals space 2.23 cross times 10 to the power of negative 2 end exponent space space mol divided by lit Amount space of space Ca left parenthesis OH right parenthesis subscript 2 space in space 500 space ml space of space saturated space solution space comma space space space space space space space space space space space space space equals space fraction numerator 2.23 cross times 10 to the power of negative 2 end exponent over denominator 2 end fraction cross times 74 space space space space space space space space space space space space space space equals 82.39 cross times 10 to the power of negative 2 end exponent space straight g space space space space space space space space space space space space space space space space equals space 823.9 space mg Now comma Volume space of space 0.4 space NaOH space equals space 500 space ml space Volume space of space Ca left parenthesis OH right parenthesis subscript 2 space space solution space taken space equals 500 space ml Volume space after space mixing space equals space 1000 space ml Molarity space of space NaOH space in space mixture space equals space fraction numerator 500 cross times 0.4 over denominator 1000 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.2 space straight M therefore space left square bracket NaOH right square bracket space equals space left square bracket OH to the power of minus right square bracket space equals space 0.2 space straight M straight K subscript sp space space equals space left square bracket Ca to the power of 2 plus end exponent right square bracket space left square bracket OH to the power of minus right square bracket space space space space space space space space space equals space 4.42 cross times 10 to the power of negative 5 end exponent space left square bracket Ca to the power of 2 plus end exponent right square bracket space equals space fraction numerator space 4.42 cross times 10 to the power of negative 5 end exponent space over denominator left square bracket OH to the power of minus right square bracket space end fraction space space space space space space space space space space space space equals space 110.5 cross times 10 to the power of negative 5 end exponent space mol divided by lit Amount space of space Ca left parenthesis OH right parenthesis subscript 2 space in space mixture space solution space equals space 110.5 cross times 10 to the power of negative 5 end exponent space cross times 74 space cross times cross times space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space space 81.8 space mg space Amount space of space Ca left parenthesis OH right parenthesis subscript 2 space preipitated space space equals space 823.9 space mg space minus space 81.8 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 742.1 space mg
 
Amount of Ca(OH)2 precipitated = 742.1 mg.
 
Answered by Varsha | 24 Dec, 2018, 01:18: PM

Application Videos

Concept Videos

Multiple Choice Questions Practice Test
Answered Unanswered My Questions My Answers
JEE main - Chemistry
number of halobenzene from the following that can be prepared by sanmeyer reaction
question image
Asked by muppanenicharitha | 14 Apr, 2024, 08:23: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Chemistry
please give me the answer of this question
question image
Asked by ruchisharmatbn | 06 Apr, 2024, 08:42: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Chemistry
distance of closest approach in chemistry
Asked by adityadoodi3 | 05 Apr, 2024, 11:27: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Chemistry
A person counts 1coin/SEC. HOW many years it will take for him to count p.5 MOLE coins
Asked by gmafia618 | 04 Apr, 2024, 08:48: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Chemistry
A solution containing 0.85 g of Z n C l 2 in 125.0 g of water freezes at − 0.23 o C . The apparent degree of dissociation of the salt is : ( k f for water= 1.86 K k g m o l − 1 , atomic mass; Z n = 65.3 and C l = 35.5 )
Asked by amarnathreddyp19 | 29 Mar, 2024, 06:47: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Chemistry
answer of 7th question
question image
Asked by syamalanandini49 | 19 Mar, 2024, 11:58: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Chemistry
jk
question image
Asked by durgeshsing666 | 18 Mar, 2024, 06:34: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Chemistry
find kc
question image
Asked by sanatkumar7799 | 13 Mar, 2024, 12:20: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Chemistry
lupac
Asked by surajchakravarti66 | 11 Mar, 2024, 11:08: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Chemistry
answer
question image
Asked by kuruvakousalya4 | 06 Mar, 2024, 10:27: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
My Class Videos Tests ₹ Plans Ask a Doubt
Get Latest Study Material for Academic year 24-25 Click here
×