Question

1. Calculate the number of moles of sulfuric acid that is contained in 250 mL of 8.500 M sulfuric acid solution

2. 7.300 moles of sodium nitrite are needed for a reaction. The solution is 5.450 M. How many mL are needed?

3. What mass (in g) of NH3 must be dissolved in 875 g of methanol to make a 0.430 molal solution?

4. Calculate the molality of a solution that is prepared by mixing 259.5 mL of CH3OH
(d = 0.792 g/mL) and 1387 mL of CH3CH2CH2OH (d = 0.811 g/mL)

5. A solution is prepared by dissolving 318.6 g sucrose (C12H22O11) in 4905 g of water. Determine the molarity of the solution

Answer 1

1. 2.125 moles

2. 1,339.45 mL

3. 6.1079 grams

4. 5.7026 M

5. 0.1898 mol/L

Step-by-step explanation:

1. Molarity is defined as the number of moles of a substance per litre solution.

-Given the volume of the solution is 250 mL and 8.5 M of sulfuric acid, the moles is obtained as:

`Moles=Concentration * Volume\\\\n=CV\\\\\\=0.25\ \ * 8.5\\\\=2.125 \ mol`

Hence, there are 2.125 moles of sulfuric acid in the solution.

2. Given 7.3 moles of sodium nitrate in a solution whose concentration is 5.45M.

#We apply the molarity formula to determine the solution's volume:

`moles=Concentration * Volume\\\\Volume=(moles)/(Concentration)\\\\=(7.3\ mol)/(5.450\ M)* 1000\ mL\\\\=1,339.45\ mL`

Hence, the volume of the solution is 1,339.45 mL

3. Given the mass of the solvent, methanol is 875 g and the molality of the solution is 0.430 m

#we apply the molality formula to calculate the mass of the solute :

`molality=(mass)/(volume)\\\\\therefore mass=volume * molality\\\\\\Molar \ mass NH_3=17.031\ g/mol\\\\=0.875* 0.430* 17.031\\\\=6.4079\ g`

Hence, the mass of the solute is 6.4079 grams

4. Given the densities as
`CH_3OH=0.792\ g/mol, \ \ CH_3CH_2CH_2OH=0.811\ g/mol`

We apply the molality formula to find the molality of the solution as follows:

`Molality, C_m=(dV_(CH_3OH))/(m_(CH_3OH)* dV_(CH_3CH_2CH_2OH))\\\\\\=(259.5\ mL* 0.792\ g/mol* 1000\ L)/(32.04\ g/mol* 0.811\ g/mol* 1387\ mL)\\\\=5.7026\ m`

Hence, the molality of the solution is 5.7026 M

5. Molarity is defined as the number of moles of a solute in a 1000 ml of a solution:

`C_m=(mass(C_(12)H_(22)O_(11)))/(Molar \ mass(C_(12)H_(22)O_(11))* mass(H_2O))\\\\\\=(318.6\ g)/(342.3\ g/mol* 4.905\ L)\\\\\\=0.1898\ mol/L`

Hence, the molality of the solution is 0.1898 mol/L