A) 1021
B) 1022
C) 1023
D) 1024
Correct Answer: C
Solution :
In 10 g of sea water, amount of NaCI = 95% of 10 g \[=\frac{95\times 10}{100}=9.5\,g\] Molecular weight of NaCI = 23 + 35.5 = 58.5 \[\because \] 58.5 g of NaCI contains \[=6.023\times {{10}^{23}}\] molecules \[\therefore \] 9.5 g of NaCI will contain \[=\frac{6.023\times {{10}^{23}}\times 9.5}{58.5}\] \[\approx 1.0\times {{10}^{23}}\]
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