JEE MAIN - Chemistry (2004 - No. 52)
6.02 $$\times$$ 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is
(Avogadro constant, NA = 6.02 $$\times$$ 1023 mol-1)
(Avogadro constant, NA = 6.02 $$\times$$ 1023 mol-1)
0.02 M
0.01 M
0.001 M
0.1 M
Объяснение
Moles (n) = $${{molecules} \over {{N_A}}}$$
$$ \Rightarrow $$ $${{6.02 \times {{10}^{20}}} \over {6.02 \times {{10}^{23}}}}$$
$$ \Rightarrow $$ 0.001
Concentration (M) = $${{moles} \over {volume}}$$
Given volume of urea is 100 ml = 0.1 litre
[Note : While calculating Concentration volume should always be in litre.]
$$ \Rightarrow $$ $${{0.001} \over {0.1}}$$
$$ \Rightarrow $$ 0.01 M
$$ \Rightarrow $$ $${{6.02 \times {{10}^{20}}} \over {6.02 \times {{10}^{23}}}}$$
$$ \Rightarrow $$ 0.001
Concentration (M) = $${{moles} \over {volume}}$$
Given volume of urea is 100 ml = 0.1 litre
[Note : While calculating Concentration volume should always be in litre.]
$$ \Rightarrow $$ $${{0.001} \over {0.1}}$$
$$ \Rightarrow $$ 0.01 M