6.02 × 1020 molecules of urea are present in 200 mL of its solution. The concentration of urea solution is (N0 = 6.02× 1023 mol-1)

Question

6.02 × 1020 molecules of urea are present in 200 mL of its solution. The concentration of urea solution is (N0 = 6.02× 1023 mol-1) (A) 0.001 M (B) 0.01 M (C) 0.02 M (D) 0.10 M2

Tardigrade Admin · Accepted Answer

Molarity (mol L-1) =( textMoles of solute/ textVolume of solution (in L)) 6.02 × 1023 molecules of urea = 1 mole ∴ 6.02 × 1020 molecules of urea =(6.02× 1020/6.02× 1023)=1× 10-3 mol 200 ml =0.20 L Molarity =(1× 10-3/0.20)=0.02 M

Q. $6.02 \times 1 0^{20}$ molecules of urea are present in 200 mL of its solution. The concentration of urea solution is $(N_{0} = 6.02 \times 1 0^{23} m o l^{- 1})$

$0.001 M$

$0.01 M$

$0.02 M$

$0.10 M 2$

Q. 6.02×1020 molecules of urea are present in 200 mL of its solution. The concentration of urea solution is (N0​=6.02×1023mol−1)

0.001M

0.01M

0.02M

0.10M2

Molarity (molL−1)=Volume of solution (in L)Moles of solute​ 6.02×1023 molecules of urea = 1 mole ∴6.02×1020 molecules of urea =6.02×10236.02×1020​=1×10−3mol 200ml=0.20L Molarity =0.201×10−3​=0.02M

Q. $6.02 \times 1 0^{20}$ molecules of urea are present in 200 mL of its solution. The concentration of urea solution is $(N_{0} = 6.02 \times 1 0^{23} m o l^{- 1})$

$0.001 M$

$0.01 M$

$0.02 M$

$0.10 M 2$