Question

Question #ec0b6

Answer 1

Calculation of Molarity

For proper definition of Molarity and other relationships refer
here .

Molarity is defined as the number of moles of solute per unit volume of the solution.

We need `2.0`M solution of `NaCl`.
That means there has to 2 moles of solute in a litre of solution. For 250 ml it is 0.5 moles. If you know the Molecular weight of `NaCl` you multiply with 0.5 to get the answer.

`Molarity = n/V`
Given
`Molarity = 2.0,V = 250 ml`
Required number of moles `= 2*0.25= 0.5` moles.
Molecular weight of `NaCl = 58.44 g`
So the weight of required `NaCl = 0.5*58.44=29.22 g`

Answer 2

Here's how you'd go about solving this problem.

You know that you must have a `"2.0 M"` `"NaCl"` solution in a final volume of `"250.0 mL"`. SInce molarity is defined as moles of solute per liter of solution, you know that you need a certain amount of moles of `"NaCl"`, your solute, to prepare this particular solution.

If you find out how many moles of solute you need, you can determine the mass in grams by using sodium chloride's molar mass, which represents how much 1 mole of `"NaCl"` weighs. So,

`C = n/V => n_("NaCl") = C * V = 2.0 "mol"/"L" * 250.0 * 10^(-3)"L"`

`n_(NaCl) = "0.5 moles"`

(notice I've converted mL to L, sicne molarity is defined as moles per liter)

This means that you need `"0.5 moles"` of sodium chloride to make a `"250.0-mL"`, `"2.0-M"` solution.

Now to determine the mass in grams

`"0.5 moles" * ("58.5 g")/("1 mole") = "29.3 g NaCl"`

As a conclusion, if you dissolve `"29.3 g"` of sodium chloride in enough water to have a volume of `"250.0 mL"`, you'll get a `"2.0 M"` solution.