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Q.

The molartiry of  HNO3  in a sample which has density 1.4 g/mL and mass percentage of 63% is ______ ( Molecular weight ofHNO3=63  )

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answer is 14.

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Detailed Solution

Mass percentage is 63% which means that 63 g of solute present in 100 g of solution.

Density=MassVolume 1.4 g/mL=100volume volume=71.428 mL

Molarity is calculated as:

molarity=massmolar mass×1000volume in mL molarity=63×100063×71.428=14 M

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The molartiry of  HNO3  in a sample which has density 1.4 g/mL and mass percentage of 63% is ______ ( Molecular weight ofHNO3=63  )