the vapour pressure of water is 92 mm. 18.1g of urea are dissolved in 100g of water. The vapour pressure is reduced by 5mm. Calculate the molar mass of urea.
Answers
Answer:
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Explanation:
As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.
The formula for relative lowering of vapor pressure will be,
\frac{p^o-p_s}{p^o}=\frac{w_2M_1}{w_1M_2}popo−ps=w1M2w2M1
where,
p-p^op−po = lowering in vapour pressure= 5 mm
p^0p0 = vapor pressure of pure solvent (water) = 92 mm
w_2w2 = mass of solute (urea) = 18.1 g
w_1w1 = mass of solvent (water) = 100 g
M_1M1 = molar mass of solvent (water) = 18 g/mole
M_2M2 = molar mass of solute (urea) = M g/mole
Now put all the given values in this formula ,we get the vapor pressure of the solution.
\frac{5}{92}=\frac{18.1 \times 18}{100\times M}925=100×M18.1×18
M=60g/molM=60g/mol
Therefore, the molar mass of urea is 60 g/mol