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CBSE Class 12-science Answered

how many grams of glucose should be dissolved in 180g of water to reduce its vapour pressure by 20%
Asked by banga71 | 17 May, 2017, 07:25: PM
answered-by-expert Expert Answer
Molar space mass space of space glucose space equals space 180.16 space straight g divided by mol straight P subscript straight s space equals space 20 percent sign space of space straight P to the power of straight o Hence comma space straight P subscript straight s space equals space 0.20 straight P to the power of straight o Solute space equals space straight w over straight M equals space fraction numerator straight w over denominator 180.16 end fraction space mol Solvent space left parenthesis water right parenthesis space equals space fraction numerator 180 space straight g space over denominator 18 space straight g space mol to the power of negative 1 end exponent end fraction space equals space 10 space moles Now comma space fraction numerator straight P to the power of straight o space minus space straight P subscript straight s over denominator straight P to the power of straight o end fraction equals straight x subscript 2 fraction numerator straight P to the power of straight o space minus space 0.20 straight P to the power of straight o over denominator straight P to the power of straight o end fraction equals fraction numerator straight n subscript 2 over denominator straight n subscript 2 plus straight n subscript 1 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 equals space fraction numerator begin display style fraction numerator straight w subscript 2 over denominator 180.16 end fraction end style over denominator begin display style fraction numerator straight w subscript 2 over denominator 180.16 end fraction plus 10 end style end fraction space equals space fraction numerator straight P to the power of straight o space minus space 0.20 straight P to the power of straight o over denominator straight P to the power of straight o end fraction space 1 minus 0.20 space equals space fraction numerator begin display style fraction numerator straight w subscript 2 over denominator 180.16 end fraction end style over denominator begin display style fraction numerator straight w subscript 2 over denominator 180.16 end fraction plus 10 end style end fraction 0.8 space equals space fraction numerator begin display style fraction numerator straight w subscript 2 over denominator 180.16 end fraction end style over denominator begin display style fraction numerator straight w subscript 2 over denominator 180.16 end fraction plus 10 end style end fraction 0.8 left parenthesis fraction numerator straight w subscript 2 over denominator 180.16 end fraction plus 10 right parenthesis space equals space space fraction numerator straight w subscript 2 over denominator 180.16 end fraction 8 space equals space fraction numerator 0.2 straight w subscript 2 over denominator 180.16 end fraction straight w subscript 2 space equals space fraction numerator 8 straight x 180.16 over denominator 0.2 end fraction equals 7206.4 space straight g space space
Answered by Prachi Sawant | 29 May, 2017, 01:06: PM
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CBSE 12-science - Chemistry
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CBSE 12-science - Chemistry
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CBSE 12-science - Chemistry
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how many ml of 0.1M HCL are required to react completely with 1g mixture of and NaHCO3 containing equimolar amounts of both?
Asked by rakeshraghav33 | 18 Sep, 2018, 04:22: PM
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