Molality
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- 1.48 M
- 7.61 M
- 5.83 M
- 9.25 M
The percentage composition of the anhydrous salt is:
Al=10.5%, K=15.1%, S=24.8% and oxygen =49.6%
[Molar mass of Al=27, K=39, S=32].
What is the empirical formula of the salt?
- K2AlS2O7
- K2Al2S2O7
- KAlS2O8
- K3AlS2O12
- 0.001 M
- 1 M
- 0.01 M
- 0.1 M
What are Molarity and molality represented by respectively?
m, M
M, m
Mo, mol
ar, al
- 20.64 m
- 13..88 m
- 18.26 m
- 18.52 m
The following table gives the solubility at different temperatures Answer the question based on this.
Temperature in Kelvin |
Substance dissolved | 283 | 293 | 313 | 333 | 353 |
KNO3 | 21 | 32 | 62 | 106 | 167 |
NaCl | 36 | 36 | 36 | 36 | 37 |
KCl | 35 | 35 | 40 | 46 | 54 |
NH4Cl | 24 | 37 | 41 | 55 | 66 |
Find the solubility of each salt at 293 K.
Percent by mass of a solute (molar mass = 28 g) in its aqueous solution is 28. Calculate the mole fraction (X) and molality (m) of the solute in the solution.
X = 0.2, m = 10
X = 0.2, m = 1259
X = 0.8, m = 1259
X = 0.8, m = 10
- 12.50 m
- 7.50 m
- 18.50 m
- 3.25 m
(Molar mass of solvent = 40 g/mol)
- 18.52 m
- 12.85 m
- 8.33 m
- 15.24 m
- 1 m
- 2 m
- 0.5 m
- 1.5 m
- 8.82 m
- 12.60 m
- 15.60 m
- 10.67 m
- 1.25 m
- 2.50 m
- 2.08 m
- 1.52 m
- 0.5
- 0.25
- 0.75
- 0.9
What is the molality of a solution having 14.2 gm of Na2SO4 dissolved in 1 liter solution?
14.2
1.42
1
cannot be determined
- 10.6 m
- 12.5 m
- 19.2 m
- 15.7 m
- 8.16 m
- 8.6 m
- 1.02 m
- 10.8 m
Identify the value of , if molal solution of a compound in benzene has mole fraction of solute equal to
- 18.428 m
- 10.428 m
- 20.428 m
- 9.428 m
- 1.79 m
- 2.143 m
- 1.951 m
- None of these
- 5.12 m
- 6.29 m
- 6.69 m
- 4.85 m
What is the mole fraction of the solute in a 1.00 m aqueous solution?
1.770
0.0354
0.0177
0.177
- 1.78 M
- 2.00 M
- 2.05 M
- 2.22 M
- 62.08 %, 7.73 M
- 37.92 %, 7.73 M
- 42.73 %, 3.86 M
- 58.58 %, 3.86 M
Percent by mass of a solute (molar mass = 28 g) in its aqueous solution is 28. Calculate the mole fraction (X) and molality (m) of the solute in the solution.
X = 0.2, m = 10
X = 0.2, m = 1259
X = 0.8, m = 1259
X = 0.8, m = 10
- 36
- 200
- 500
- 18
(Density of the solution is 1.10 g/mL)
- 1.53 M, 1.64 m
- 1.68 M, 1.68 m
- 1.80 M 1.68 m
- 1.68 M 1.80 m
- 9.00×10−3
- 1.50×10−2
- 5.10×10−3
- 11.2×10−3
- 0.8
- 0.4
- 0.2
- 0.36
- 4.13 m
- 2.57 m
- 3.85 m
- 5.46 m