The molarity of the aqueous solution of NaNO3 is 2.30 M .
Therefore, to calculate how many grams you obtain in that particular volume, utilize the solution's density.
1.00L x 1000mL/1L x 1.153g/1mL
= 1153g
Here, indicates this solution has a 17.0% NaNO3 content by mass. This indicates that for every 100.0 g of solution, you receive 17.0 g of NaNO3.
1153g solution x 17.0g NaNO3/100g solution
= 196.01 g NaNO3
Use the molar mass of NaNO3 to calculate how many moles are contained in that many grams.
196.01 x 1 mole NaNO3/84.9947 g/mol
= 2.30 moles of NaNO3
To find molarity the formula is ,
c = n/v
= 2.30 moles / 1.00L
= 2.30 M
The molarity of NaNO3 is 2.30M.
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