Answer:
A Glucose C6H12O6 is a covalent compound while NaCl sodium chloride is a ionic compound.
B Stronger molecular forces are present in NaCl because ionic bond is stronger than covalent bond.
C melting point allow us to predict the strength of intermolecular forces
High melting point stronger molecular forces and vice versa.
Explanation:
Answer:
120.0 mL.
Explanation:
- As it is known that the no. of millimoles of a solution before dilution is equal to the no. of millimoles after dilution.
We suppose that the initial W% of methanol is 100.0 %
<em>∴ (W%V) before dilution = (W%V) after dilution.</em>
W% before dilution = 100.0 %, V before dilution = 18.0 mL.
W% after dilution = 15.0 %, V after dilution = ??? mL.
<em>∴ V after dilution = (W%V) before dilution/W% after dilution = </em>(100.0 %)(18.0 mL)/(15.0%)<em> = 120.0 mL.</em>
Answer:
2,347.8 grams
Explanation:
The freezing point depression Kf of water = 1.86° C / molal.
To still freeze at -12° C, then the molality of the solution 12/ 1.86 = 6.45 moles
The molecular weight of sorbitol (C6H14O6)is:
6 C = 6 ×12 = 72
14 H = 14 × 1 = 14
6 O = 6 × 16 = 96
...giving a total of 182
So one mole of sorbitol has a mass of 182 grams.
Since there are 2 kg of water, 2 × 6.45 moles = 12.9 moles can be added to the water to get the 12° C freezing point depression.
Therefore
grams = moles × molar mass
12.9 moles × 182 grams / mole = 2,347.8 grams of sorbitol can be added and still freeze
Answer:
34.23 g.
Explanation:
<em>Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.</em>
M = (no. of moles of solute)/(V of the solution (L)).
∴ M = (mass/molar mass)of C₁₂H₂₂O₁₁/(V of the solution (L)).
<em>∴ mass of C₁₂H₂₂O₁₁ = (M)(molar mass)(V of the solution (L)</em> = (1.0 M)((342.3 g/mol)/(0.10 L) = <em>34.23 g.</em>
