Question

30gm of urea is dissolve of 44.5gm of H₂O determine 1) %w/w 2)mole fraction of each component ​

Answer 1

Answer:-

Given:-

Mass of urea = 30 gm

Molar mass of urea (NH₂CONH₂) = 60 g/mol

Mass of H₂O = 44.5 gm

Molar mass of H₂O = 18 g/mol

We have to find:-

1. w/w %
2. Mole fraction of each component.

1. We know that,

Mass % (w/w %) = Mass of one component × 100/Total mass of all components

So,

★ (w/w) % of urea = (30 × 100) / (30 + 44.5)

⟹ (w/w) % of urea = 3000/74.5

⟹ (w/w) % of urea = 40.3 %

★ (w/w) % of water = (44.5 × 100) / (74.5)

⟹ (w/w) % of water = 4450/74.5

⟹ (w/w) % of water = 59.7 %

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2. We know,

$\sf Mole \: fraction \: of \: a \: component( X_{A}) = \dfrac{ n_{A} }{n_{A} +n_{B} }$

where; A , B are the components of a solution.

• No. of moles (n) = Given mass/Molar mass

So,

⟹ No. of moles of urea $\sf n_{urea}$ = 30/60 = 0.5 moles.

⟹ No. of moles of water $\sf n_{water}$ = 44.5/18 = 2.4 moles

Therefore,

$\sf \: X_{urea} = \dfrac{n_{urea}}{n_{urea} + n_{water}} \\ \\ \\ \implies \sf \: X_{urea} = \frac{0.5}{0.5 + 2.4} \\ \\ \\ \sf \sf \: X_{urea} = \frac{0.5}{2.9} \\ \\ \\ \implies \boxed{\sf \: X_{urea} =0.18}$

Similarly,

$\sf \: X_{water} = \dfrac{n_{water}}{n_{urea} + n_{water}} \\ \\ \\ \implies \sf \: X_{water} = \frac{2.4}{2.9} \\ \\ \\ \implies \boxed{\sf \: X_{water} =0.82}$