Question

How many grams of solute are contained in a given volume of solution: a) 200 ml of a 0.3 M glucose

solution; b) 400 ml of 0.15 n NaCl solution
​

Answer 1

Answer:

Explanation:

Let's start off with the equation for molarity

1000

m

L

=

1

L

Therefore, if we divide 50mL by 1000mL we will obtain a value of 0.05L.

Next, the equation has to be rearranged to solve for the moles of solute:

Moles of solute = Molarity  

×

Liters of solution

Now, multiply 0.360 M by 0.05:

0.360 mol

1 L

×

0.05 L

=

0.018 mol

To obtain the mass of solute, we will need to the molar mass of NaCl, which is 58.44 g/mol:

Finally, multiply the number of moles by 58.44 g/mol

0.018

mol

×

58.44

g

1

mol

Boom, here it is:

1.05

g

Answer 2

There are 1.05g of solute present.

Explanation:

Let's start off with the equation for molarity:

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:

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

1000

m

L

=

1

L

Therefore, if we divide 50mL by 1000mL we will obtain a value of 0.05L.

Next, the equation has to be rearranged to solve for the moles of solute:

Moles of solute = Molarity

×

Liters of solution

Now, multiply 0.360 M by 0.05:

0.360 mol

1 L

×

0.05 L

=

0.018 mol

To obtain the mass of solute, we will need to the molar mass of NaCl, which is 58.44 g/mol:

Finally, multiply the number of moles by 58.44 g/mol

0.018

mol

×

58.44

g

1

mol

Boom, here it is:

1.05

g