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CBSE Class 12-science Answered

12gram glucose ko 10 gm solution Ka kwathnank 100.34 c hai,molal unnayan sthirank batao  
Asked by mishrasatyamcs12 | 17 Sep, 2019, 03:15: AM
answered-by-expert Expert Answer
12 gram of glucose is present in 100 gram  of solution and boining point of solution is 100.34 o C.  Molal elevation constant=?
Elevation in boiling point can be determined by formula-
increment T subscript b space equals m cross times k subscript b increment T subscript b space equals fraction numerator W e i g h t space o f space s o l u t e cross times 1000 over denominator M o l k e c u l a r space m a s s space o f space s o l u t e space cross times thin space W e i g h t space o f space s o l v e n t end fraction cross times k subscript b 100.34 minus 100 equals fraction numerator 12 cross times 1000 over denominator 180 cross times left parenthesis 100 minus 12 right parenthesis end fraction cross times k subscript b .34 equals fraction numerator 12 cross times 1000 over denominator 180 cross times 88 end fraction cross times straight k subscript straight b straight k subscript straight b equals 0.4488 space K space k g space m o l to the power of negative 1 end exponent
Answered by Ravi | 17 Sep, 2019, 12:03: PM

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