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

Step-by-step explanation:

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

`Molarity=(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=(xg)/(85g/mol)`

`x=238g`

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