Calculate the molarity of pure water using its density to be $1000 k g m^{- 3}$ .

66.63 M

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46.92 M

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55.55 M

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Solution

The correct option is C 55.55 M
Molarity= $\frac{M o l e s o f s o l u t e}{V o l u m e o f s o l u t i o n i n l i t r e s}$
= $\frac{\frac{1000}{18} \times 10^{3}}{{1 m}^{3} \times 1000 (L)}$
= $55.55 M$

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Calculate the molarity of pure water using its density to be 1000kgm−3.

The correct option is C 55.55 MMolarity= MolesofsoluteVolumeofsolutioninlitres =100018×1031m3×1000(L) =55.55M

Calculate the molarity of pure water using its density to be $1000 k g m^{- 3}$ .

The correct option is C 55.55 M
Molarity= $\frac{M o l e s o f s o l u t e}{V o l u m e o f s o l u t i o n i n l i t r e s}$
= $\frac{\frac{1000}{18} \times 10^{3}}{{1 m}^{3} \times 1000 (L)}$
= $55.55 M$