Question

What mass of `"ammonium chloride"` solute is required to prepare a `200*mL` of solution for which the concentration was `0.100*mol*L^-1`?

Answer 1

`1.07g`

Explanation:

First calculate the mass of one mole of `NH_4Cl`. This would produce a `1.00 (mol)/L` solution in 1 Liter. Then take `1/10` of that to make `0.100` mol, and then `1/5` of that for the `200.00` mL solution (`200/1000`).

M.W. `NH_4Cl` = `14 + 4 + 35.5 = 53.5g`

`53.5 xx 1/10 xx 1/5 = 1.07g` in `200.00` mL solvent.

Answer 2

Approx. `1*g` solute are required....

Explanation:

To define the `"concentration"` we use the quotient....

`"concentration"="Moles of solute"/"Volume of solution"`

And thus........

`"moles of ammonium chloride"="volume"xx"concentration"`

`=200.0*cancel(mL)xx10^-3*cancelL*cancel(mL^-1)xx0.100*mol*cancel(L^-1)`

`=0.020*mol`

And we take the product of this molar quantity by the molar mass of ammonium chloride to get the required mass of solute....

`=0.020*molxx53.49*g*mol^-1=??*g`.

Answer 3

`"1.0698 g"`

Explanation:

Moles of solute`= "0.100 mol"/cancel"L" × 200.00 cancel"ml" × (10^-3 cancel"L")/cancel"ml"= "0.02 mol"`

Molar mass of `"NH"_4"Cl"` is `"53.491 g/mol"`

Mass of solute`= 0.02 cancel"mol" × "53.491 g"/cancel"mol" = "1.0698 g"`