JEE Main & Advanced Sample Paper JEE Main Sample Paper-9

  • question_answer
    Vapour pressure of a solution of 5 g of non - electrolyte in 100 g of water at a particular temperature is \[2985\text{ }N/{{m}^{2}}.\] The vapour pressure of pure water is \[3000\text{ }N/{{m}^{2}},\]the molecular weight of the solute is

    A)  90                         

    B)  200    

    C)  180                       

    D)  380

    Correct Answer: C

    Solution :

    \[{{M}_{2}}=\frac{{{w}_{2}}{{M}_{1}}\times {{p}^{0}}}{{{w}_{1}}({{p}^{0}}-{{p}_{s}})}\] Where, \[{{M}_{2}}=\]molar mass of solute \[{{w}_{2}}=\]weight of solute \[{{M}_{1}}=\] molar mass of solvent \[{{w}_{1}}=\]weight of solvent \[{{p}^{0}}=\]vapour pressure of pure solvent \[{{p}_{s}}=\]vapour pressure of solution \[=\frac{5\times 18\times 3000}{100\times (3000-2985)}=180\]


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