1.

6.02 xx 10^(20) molecules of urea are present in 100 mL of its solution. The concentration of urea solution is (Avogadro constant, N_(A) = 6.02 xx 10^(23)mol^(-1))

Answer»

0.02 M
0.01 M
0.001 M
0.1 M Glucose.

Solution :`"No. of MOLES of urea"=(6.02xx10^(20))/(6.02xx10^(23))`
=0.001 mol.
100 mL of solution has moles = 0.001
1000 mL of solution has moles
`=(0.001)/((10ML))XX(1000mL)=0.01 M.`


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