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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

Respuesta :

opudodennis

Answer:

1.    2.125 moles

2.   1,339.45 mL

3.   6.1079 grams

4.   5.7026 M

5.   0.1898 mol/L

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:

[tex]Moles=Concentration \times Volume\\\\n=CV\\\\\\=0.25\ \ \times 8.5\\\\=2.125 \ mol[/tex]

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:

[tex]moles=Concentration \times Volume\\\\Volume=\frac{moles}{Concentration}\\\\=\frac{7.3\ mol}{5.450\ M}\times 1000\ mL\\\\=1,339.45\ mL[/tex]

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 :

[tex]molality=\frac{mass}{volume}\\\\\therefore mass=volume \times molality\\\\\\Molar \ mass NH_3=17.031\ g/mol\\\\=0.875\times 0.430\times 17.031\\\\=6.4079\ g[/tex]

Hence, the mass of the solute is 6.4079 grams

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

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

[tex]Molality, C_m=\frac{dV_{CH_3OH}}{m_{CH_3OH}\times dV_{CH_3CH_2CH_2OH}}\\\\\\=\frac{259.5\ mL\times 0.792\ g/mol\times 1000\ L}{32.04\ g/mol\times 0.811\ g/mol\times 1387\ mL}\\\\=5.7026\ m[/tex]

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:

[tex]C_m=\frac{mass(C_{12}H_{22}O_{11})}{Molar \ mass(C_{12}H_{22}O_{11})\times mass(H_2O)}\\\\\\=\frac{318.6\ g}{342.3\ g/mol\times 4.905\ L}\\\\\\=0.1898\ mol/L[/tex]

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