1.2 gm of non volatile substance was dissolved in 100gm of acetone at 20.c , the vapor pressure of solution was found to be 182.5 torr
Find the molar mass of the substance. Given v.p of acetone at 20.c =185 torr
Answers
Answer:
yoyo
Explanation:
sup
Answer:
po = 185 torr at 20oC
ps = 183 torr at 20oC
Mass of non-volatile substance,
m= 1.2 g
Mass of acetone taken = 100 g
As we have,fraction numerator straight p subscript straight o minus straight p subscript straight s over denominator straight p subscript straight s end fraction space equals space straight n over straight N
putting the values, we get
fraction numerator 185 minus 183 over denominator 183 end fraction space equals space fraction numerator begin display style fraction numerator 1.2 over denominator straight M end fraction end style over denominator begin display style 100 over 58 end style end fraction space
rightwards double arrow 2 over 183 space equals space fraction numerator 1.2 space straight x space 58 over denominator 100 space straight x space straight M end fraction
therefore space straight M space equals space fraction numerator 183 space straight x space 12 space straight x space 58 over denominator 2 space straight x space 100 end fraction
straight M space equals space 63.684 space almost equal to space 64 space straight g divided by mol
Given,
po = 185 torr at 20oC
ps = 183 torr at 20oC
Mass of non-volatile substance,
m= 1.2 g
Mass of acetone taken = 100 g
As we have,fraction numerator straight p subscript straight o minus straight p subscript straight s over denominator straight p subscript straight s end fraction space equals space straight n over straight N
putting the values, we get
fraction numerator 185 minus 183 over denominator 183 end fraction space equals space fraction numerator begin display style fraction numerator 1.2 over denominator straight M end fraction end style over denominator begin display style 100 over 58 end style end fraction space
rightwards double arrow 2 over 183 space equals space fraction numerator 1.2 space straight x space 58 over denominator 100 space straight x space straight M end fraction
therefore space straight M space equals space fraction numerator 183 space straight x space 12 space straight x space 58 over denominator 2 space straight x space 100 end fraction
straight M space equals space 63.684 space almost equal to space 64 space straight g divided by mol
Explanation:
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