Question

how many grams of sodium nitrate are needed to make 2.50L of 1.12m solution ​

Answer 1

Answer:  238 g of sodium nitrate are needed to make 2.50L of 1.12m solution

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Now put all the given values in the formula of molality, we get

$1.12=\frac{\text {moles of}NaNO_3}{2.50}$

moles of $NaNO_3$ = 2.8

moles of $NaNO_3=\frac{\text {given mass}}{\text {Molar mass}}$

$2.8mol=\frac{xg}{85g/mol}$

$x=238g$

Thus 238 g of sodium nitrate are needed to make 2.50L of 1.12m solution ​