Question
Question: How much urea (molar mass \[60g/mol\]) should be dissolved in \[50g\] of water so that its vapour pr...
How much urea (molar mass 60g/mol) should be dissolved in 50g of water so that its vapour pressure at room temperature is reduced by 25%. [Write up to 1 decimal place]
Solution
As we know that the urea is a chemical compound having the chemical formula, CO(NH2)2. The urea contains carbon, nitrogen and oxygen atoms. And the urea is also known as carbamide which contains two amino groups. This amino group is joined with the carbonyl functional group. The molar mass of urea is equal to 60g/mol. And water is a chemical compound having the symbol H2O with molecular mass 18g.
Complete answer:
The weight of the urea can be find out by using the equation,
P10P10−P1=W2×M1W1×M2 _____ (1)
Where, P1o is the vapour pressure of the pure solvent, X.
P1is equal to the vapour pressure of the solution of urea. Which implies,
P1o−P1=25% of x=18.02g/mol and W1is equal to the weight of water which is equal to 50g. M2is equal to the molar mass of urea which is equal to 60g/mol.
Substitute the values in the equation one, will get the weight of phenol dissolved in 50g of water;
x0.25x=50×60W2×18.02
By rearranging the equation,
W2=18.020.25×3000=41.62g≃41.7g
Therefore, the weight of phenol dissolved in 50g of water is equal to 41.7g.
Note:
We have to remember that the elevation in boiling point explains the method that the boiling point of a liquid will be more by adding another compound. And the vapour pressure is also known as equilibrium vapour pressure which is equal to the exerted pressure by a vapour in the equilibrium in thermodynamics in a closed system with its condensed phases at a particular temperature. And the weight of the solution can be found by using the equation of vapour pressure.