Question

Mole fractions of A and B in a mixture in which $6.023\times {10}^{23}$ molecules of A and $10.4\times {10}^{23}$ molecules of B are present.

• A. 0.37, 0.37
• B. 063, 0.37
• C. 0.37, 0.63
• D. 0.63, 0.63

Answer 1

The correct option is C

0.37, 0.63

To find mole fraction, we need number of moles.

$N_A$ = Number of moles of A = $\frac{6.023\times {10}^{23}}{6.023\times {19}^{23}}=1$

$N_B$ = Number of moles of B = $\frac{10.4\times {10}^{23}}{6.023\times {10}^{23}}=1.73$

$X_A=\frac{N_A}{N_A+N_R}=\frac{1}{2.73}=0.37$

$X_B=\frac{N_B}{N_A+N_R}=\frac{1.73}{2.73}=0.63$