Question

How many mole of Urea should required to prepare 1.25m,250ml of solution?​

Answer 1

3.60g of NaOH is required to prepare that solution

We are given the molarity and the volume of solution. The only issue is that the volume is given in mL instead of L. This issue can be fixed by using the following conversion factor:

1000

m

L

=

1

L

Therefore, if we divide 200mL by 1000mL we will obtain a value of 0.200 L.

Rearrange the equation to solve for moles of solute:

Moles of solute = Molarity

×

Liters of solution

Multiply 0.450 M by 0.200:

0.450 mol

1 L

×

0.200 L

=

0.09 mol

To obtain the mass of solute, we will need to the molar mass of NaOH, which is 40.00 g/mol:

Finally, multiply the number of moles by 40.00 g/mol

0.09

mol

×

40.00

g

1

mol

=

3.60

g

Boom, you have a mass of:

3.60

g