What is the molality of an aqueous solution that is 15.3 ppm MnSO_4? The density

Question

What is the molality of an aqueous solution that is 15.3 ppm MnSO_4? The density of this solution is 1.12 g/mL 20.0 g of an unknown nonvolatile molecular solid was added to 125 g H_2 O at 25.0 degree C. The vapor pressure of this solution was found to be 22.67 torr. If the vapor pressure of pure water at 25.0 degree C is 22.76 torr, what is the molar mass of the unknown solid? What mass of NaCl must be added to 100.0 mL H_2O to create a solution that is isotonic with blood at 25.0 degree C? (blood has an osmotic pressure of 7.70 atm at 25.0 degree C) Which of the following reactions is predicted to have a negative Delta S a. NaCl(s) rightarrow Na^+(aq) + Cl^- (aq) C_2 H_2(g) + 4F_2 (g) rightarrow 2CF_4(g) + H_2(g) The melting point of Tungsten is 3422 degree C. If Delta H_fus = 35.2 kJ/mol, what is the entropy of fusion? At what temperature is the following reaction spontaneous? 2 SO_3 (g) rightarrow 2SO_2 (g) + O-2 (g)

Explanation / Answer

4. 15.3 ppm = 15.3mg MnSO4 / L of solution = 0.0153 g/L

2) Divide by the molecular weight for MnSO4:

0.0153 g/L divided by 151g/mol = 1.01 * 10-4 mol/L = 1.01 * 10-4 M

First let's figure out the mass of the solution using dimensional analysis and density:
? g soln = 1.00 L x (1000 mL)/1 L x (1.12 g)/1 mL = 1120 g solution

Now we need to know the mass of just the water.
? g water = mass of the solution - mass of MnSO4 in the solution
= 1120 g solution - (1.01 * 10-4 mol MnSO4 x 151 g/mol)
= 1120 g solution - 0.015 g MnSO4
= 1119 g water or 1.119 kg of water

Now, use the units of molality as an equation: the molality (mol solute/kg solvent) = (no. moles MnSO4)/(no. kg water)

So, molality = (1.01 * 10-4 mol MnSO4)/(1.119 kg water) = 9.02 * 10-5m MnSO4

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