Vapour pressure of water is 360 …

Vapour pressure of pure water, p^o = 360 mm Hg

Lowered vapour pressure, p = 360 –0 .5% of 360

= 360 – 1.8 = 358.2 mm Hg

Now, Weight of water, w1= 200ml x 1g.ml = 200g

Weight of urea w2 = ?

Molecular weight of water, M1 = 18 g/mol Molecular weight of urea, M2 = 60 g/mol

According to Roult’s law:

{tex}{p^o -p \over p} = {n_2\over n_1 + n_2}{/tex}

=> {tex}{360 - 358.2\over 358.2} = { {w_2 \over M_2}\over {{w_1\over M_1 } + {w_2 \over M_2}}}{/tex}

=> {tex}0.005 = { {w_2 \over M_2}\over {{w_1\over M_1 } + {w_2 \over M_2}}}{/tex}

=> {tex}0.005 {\left ({w_1\over M_1 } + {w_2 \over M_2}\right )}= {w_2 \over M_2}{/tex}

=> {tex}0.005 \times {w_1\over M_1 } = {w_2 \over M_2} - 0.005\times {w_2 \over M_2}{/tex}

=> {tex}0.005 \times {w_1\over M_1 } =0.995\times {w_2 \over M_2}{/tex}

=> {tex}{0.005 \over 0.995} \times {w_1\times M_2\over M_1 } = {w_2 }{/tex}

=> {tex}{0.005 \over 0.995} \times {200\over 18 }\times 60 = {w_2 }{/tex}

=> w₂ =  3.33g

Hence, 3.33g urea should be added