5000 parts of municipal water contains 1 part of urea by weight. Find the no. of molecules of urea present in 1 ml of municipal water (NA=6×1023), density of H2O is 1gm/cc ( mol wt of urea=60)

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5000 parts of municipal water contains 1 part of urea by weight. Find the no. of molecules of urea present in 1 ml of municipal water $(N_{A} = 6 \times 10^{23})$ , density of $H_{2} O$ is 1gm/cc ( mol wt of urea=60)

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$2 \times 10^{18}$

$2 \times 10^{19}$

$2 \times 10^{20}$

$5 \times 10^{18}$

answer is A.

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Detailed Solution

5000 ------------ 1 gm urea
1 gm ------------ $\frac{1}{5000} g m$
No of molecules = $\frac{1}{5000} \times \frac{1}{6010} \times 6 \times 10^{23} = \frac{1}{5} \times 10^{19} = 2 \times 10^{18}$