first determine the number ofmoles dissolved in given solution
then .5 moles moles dissolved in 800g. as comparison with 1000g of water, we know 100g of water dissolve only.1 moles of a glucose so we .7moles of glucose dissolve in 800g.
10.27
The higher the molality the higher the boiling point and the lower the freezing point. Conversely, the lower the molality the lower the boiling point and the lower the freezing point. Now let's analyze the problem: The total molality of 0.10 m of NaCl is 0.20 m. This is based on the property of the ionic bond. Ionic bonds are salts and thus strong electrolytes, which means they break up into all of there constituents in solution. Since two substances make up sodium chloride (sodium and chlorine), that means you'll have two substances overall. Your total molality was .10, and you have two substances, which means you're left with 0.10 * 2 = 0.20 m. Glucose is made of covalent bonds which do NOT break up in solution. This means you are left with one substance, C6H12O6. So you have 0.10 * 1 = 0.10. You are left comparing 0.20m of NaCL and 0.10m of C6H12O6. The higher molality has the higher boiling point.
Salt is an ionic compound, it forms ions when dissolved in water. An ionic solution conducts electricity; ammonia or glucose dissolved in water will not conduct electricity as they are molecules not ions. Table salt is an ionic compound, NaCl (Sodium ion and Chloride ion)
Boiling and freezing points are colligative properties, meaning they depend on the number of solute particles dissolve in solution. Glucose is a molecular compound so it is one particle dissolved in solution. CaCl2 will dissociate into three particles in solution. There are three times as many particles present in solution when CaCl2 dissolves.
A 1.0 molar solution would be made up by dissolving the the molecular weight of glucose (in grams) in one litre. The molecular weight of glucose is 180.1559g; so a 1.0M solution would be 180.1559g in 1 litre. A 0.5M solution would be half that strong, so that would need 90.0780g in 1 litre. For 500ml you'd need take half the weight into half the volume, so that would be 45.0390g dissolved in water and made up to 500ml.
i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2 i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2 i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2 i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2
This molality is 90,08 g/kg.
Assuming that is is a solution of glucose in water, the answer is 93%.
1M glucose means that 1 mole of glucose is dissolved in 1kg of water. Since 1M means 1 molal. And molality is equla to no.of moles of solute per kg of water.
3mol/6kg
3mol/6kg
4 mol over 0.800 kg
yes
Molality (m) is moles of solute per kg of solvent. 3 moles/6 kg = 0.5 molal.
If 180g glucose is present in one litre of solution then boiling point is 100.52 Celsius.
There is twice the change in colligative properties in the sodium chloride solution than in the glucose solution.
The higher the molality the higher the boiling point and the lower the freezing point. Conversely, the lower the molality the lower the boiling point and the lower the freezing point. Now let's analyze the problem: The total molality of 0.10 m of NaCl is 0.20 m. This is based on the property of the ionic bond. Ionic bonds are salts and thus strong electrolytes, which means they break up into all of there constituents in solution. Since two substances make up sodium chloride (sodium and chlorine), that means you'll have two substances overall. Your total molality was .10, and you have two substances, which means you're left with 0.10 * 2 = 0.20 m. Glucose is made of covalent bonds which do NOT break up in solution. This means you are left with one substance, C6H12O6. So you have 0.10 * 1 = 0.10. You are left comparing 0.20m of NaCL and 0.10m of C6H12O6. The higher molality has the higher boiling point.
Molarity = moles of solute/Liters of solution Find moles glucose, which is molecular formula------C6H12O6 154 grams C6H12O6 (1 mole C6H12O6/180.156 grams) = 0.8548 moles C6H12O6 Molarity = 0.8548 moles C6H12O6/1 Liter = 0.855 M glucose ---------------------------