A 0.148-M solution of X (Mm = 177 g/mol) has a density of 1.156 g/ml. What...

A 0.148-M solution of X (Mm = 177 g/mol) has a density of 1.156 g/ml. What is the molality of X in the solution?

Solutions

Expert Solution

Molarity = 0.148 = (solute weightgr/mol.wt)(1/1L)
we have mol.wt = 177g/mol

Therefore, solute weight= molarity x molecular weight
                       = 0.148 x 177
                       = 26.196 g

Weight of 1000 ml of solute = 1000 mL x density
                           = 1000 mL x 1.156 g/mL
                           = 1156 g
weight of the solvent in 1156 g of solution = 1156 g - 26.196 g
                                           = 1129.8 g



Molality = no. moles of solute present in 1 kg (or 1000 g) of the solvent.

therefore, molality = (Weight of Solute/Mol. weight)(1000/solvent weight in grams)
                   = (26.196 g/177 g/mol)(1000/1129.8 g)
                   = 0.131 m

Molality of the solution = 0.131 m

queen_honey_blossom answered 1 year ago

Next > < Previous

Related Solutions

A solution is 3.24 M NH3 (MM 17.04 g/mol) in water (MM 18.02 g/mol). The density...

A solution is 3.24 M NH3 (MM 17.04 g/mol) in water (MM 18.02 g/mol). The density of the resulting solution is 0.974 g/mL. What is the mole fraction, mass percent and molality of NH3 in the solution?

An aqueous solution is 2.00 M urea. The density of the solution is 1.029 g/ml. What...

An aqueous solution is 2.00 M urea. The density of the solution is 1.029 g/ml. What is the molality concentration of urea in the solution?

(a) 3.25 M NaCl solution (density of solution = 1.08 g/mL)

Calculate the molality of each of the following aqueous solutions. (a) 3.25 M NaCl solution (density of solution = 1.08 g/mL) m (b) 51.4 percent by mass KBr solution m

When 150.0 mL of a 0.200 M Ca(NO3)2 (MM = 164.088 g/mol) solution is added to...

When 150.0 mL of a 0.200 M Ca(NO3)2 (MM = 164.088 g/mol) solution is added to 200.0 mL of a 0.150 M K3PO4 (MM = 212.27 g/mol) solution, what mass of calcium phosphate (MM = 310.17 g/mol) will precipitate? a. 9.31 g b. 3.10 g c. 2.91 g d. 1.40 g e. 4.65 g

A 500 mL saturated solution of MgCO3 (MM = 84.3 g/mol) is reduced to 120 mL...

A 500 mL saturated solution of MgCO3 (MM = 84.3 g/mol) is reduced to 120 mL by evaporation. What mass of solid MgCO3 is formed? Ksp = 4.0 x 10-5 for MgCO3. EDIT: If ions can evaporate in water then the answer below is right. If not, this is how I solved the problem. MgCO3 dissolves at a 1:1 ratio, therefore Ksp=s2 and s=Ksp1/2 (or sqrt of Ksp) s = (4.0x10-5)1/2 = .00632 M Using the equation M1V1=M2V2 M1 =...

A solution is 5.66 % by mass NH3 (MM 17.04 g/mol) in water (MM 18.02 g/mol)....

A solution is 5.66 % by mass NH3 (MM 17.04 g/mol) in water (MM 18.02 g/mol). The density of the resulting solution is 0.974 g/mL. What is the mole fraction, molarity and molality of NH3 in the solution?

A 5.85% (w/w) Fe(NO3)3 (241.86 g/mol) solution has a density of 1.059 g/mL. Calculate

A 5.85% (w/w) Fe(NO3)3 (241.86 g/mol) solution has a density of 1.059 g/mL. Calculate(a) The molar analytical concentration of Fe(NO3)3 in this solution.(b) The molar NO3 - concentration in the solution.(c) The mass in grams of Fe(NO3)3 contained in each liter of this solution.

Question