A) 150g
B) 75g
C) 37.5g
D) 50g
Correct Answer: A
Solution :
Octane has molar mass of 114 g/mol. \[\frac{\Delta P}{P}=\frac{\frac{{{W}_{2}}}{{{M}_{2}}}}{\frac{{{W}_{2}}}{{{M}_{2}}}+\frac{{{W}_{1}}}{{{M}_{1}}}}\] \[\frac{75}{100}=\frac{\frac{{{W}_{2}}}{50g/mol}}{\frac{{{W}_{2}}}{50g/mol}+\frac{114g}{114g/mol}}\] \[0.75=\frac{\frac{{{W}_{2}}}{50}}{\frac{{{W}_{2}}}{50}+1}\] \[\frac{{{W}_{2}}}{50}+1=\frac{{{W}_{2}}}{50\times 0.75}\] \[{{W}_{2}}=150\,g\] \[{{W}_{2}}=150\,g\] Note: \[{{W}_{2}}\]and \[{{M}_{2}}\] are mass and molar mass of solute\[{{W}_{1}}\]and\[{{M}_{1}}\]and are mass and molar mass of octane.You need to login to perform this action.
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