6.02 × 1020 molecules of urea are present in 100 mL of its solution. The concentration of solution is :-

Question

6.02 × 1020 molecules of urea are present in 100 mL of its solution. The concentration of solution is :- (A) 0.1 M (B) 0.02 M (C) 0.01 M (D) 0.001 M

Tardigrade Admin · Accepted Answer

Number of moles =( text number of molecules / text NA ) =(6.02 × 1020/6.02 × 1023)=10-3 mol Molar conc =( n × 1000/ V text solution ( mL ))=(10-3 × 1000/100) Molar conc. =0.01 M

Q. $6.02 \times 1 0^{20}$ molecules of urea are present in 100 mL of its solution. The concentration of solution is :-

0.1 M

0.02 M

0.01 M

0.001 M

Number of moles $= \frac{number of molecules}{NA}$ $= \frac{6.02 \times 1 0 ^{20}}{6.02 \times 1 0 ^{23}} = 1 0^{- 3} m o l$
Molar conc $= \frac{n \times 1000}{V _{solution} ( m L )} = \frac{1 0 ^{- 3} \times 1000}{100}$
Molar conc. $= 0.01 M$

Q. 6.02×1020 molecules of urea are present in 100 mL of its solution. The concentration of solution is :-