CBSE Class 11-science Answered

Commercially available conc. HCl contains 38% HCl by mass. a) What is the molarity of this solution, if the density is 1.10g/ml? b) What volume of conc. HCl is required to make 1.0 L of 0.10M HCl?

Asked by Akanksha Srikanth | 07 Dec, 2015, 06:30: PM

Expert Answer

Hi,

$begin mathsize 11px style left parenthesis straight a right parenthesis space 38 percent sign space solution comma space mean space 38 space straight g space of space HCl space in space 100 space straight g space of space solution. Then comma space Mass space of space the space solution space equals space 100 straight g therefore space Volume space of space the space solution space equals space fraction numerator 100 space straight g over denominator Density end fraction space equals space fraction numerator 100 space straight g over denominator 1.19 space straight g divided by mL end fraction space equals space 84.0 space mL space equals space fraction numerator 84.0 over denominator 1000 end fraction space straight L Molar space mass space of space HCl space equals space 36.5 space gmol minus 1 Number space of space moles space of space HCl space dissolved space equals space 38 over 365 space mol space equals space 1.04 space mol therefore space Molarity space of space HCl space solution space equals space fraction numerator 1.04 space mol over denominator 84.1000 space straight L end fraction space equals space bold 12 bold. bold 4 bold space bold mol bold space bold L to the power of bold minus bold 1 end exponent left parenthesis straight b right parenthesis space Molarity space of space conc. space HCl space sample space equals space 12.38 space mol divided by straight L Molarity space of space HCl space solution space to space be space prepared space equals space 0.10 space mol divided by straight L Volume space of space HCl space solution space to space be space prepared space equals space 1.00 space straight L space equals space 1000 space mL By space using space molarity space equation comma straight M 1 straight V 1 space equals space straight M 2 straight V 2 VHCl space equals space fraction numerator 0.10 space cross times space 1000 over denominator 12.38 end fraction space mL space equals space bold 8 bold. bold 07 bold space bold mL Thus comma space to space obtain space 1.0 straight L space of space 0.10 space straight M space HCl comma space one space should space dissolve space 8.08 space mL space of space concentrated space HCl space to space make space the space volume space to space 1.0 straight L. end style$