Question #273023

a solution that contains 70g of NaNO3 at 30 degrees celicius in 100 ml H2O


1
Expert's answer
2021-11-30T05:00:01-0500


Molar mass of NaNO3 = 84.9947 g/mol


Moles =7084.9947=0.824moles=\frac{70}{84.9947}=0.824moles


Molarity =0.8240.1=0.0824M= \frac{0.824}{0.1}=0.0824M


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