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.
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.
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.
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
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)
Ok so all I have found so far is In Benedict's solution Starch-no observable reaction and Starch Amylase-clear blue. In Iodine Starch- Blue Black Starch and Amylase-orange the web page I found this on how ever was very confusing so I'm not sure
3mol/6kg
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.
To find the molality of a solution, you need the mass of the solvent (usually water) in kilograms and the number of moles of solute (glucose). Given that the solution is 7.80% glucose by weight, you can calculate the mass of glucose in the solution and then convert it to moles using the molar mass of glucose. From there, you can find the molality by dividing the moles of glucose by the mass of the solvent in kilograms.
The molality of the solution is calculated by dividing the moles of solute by the mass of the solvent in kg. In this case, there are 3 mol of glucose in 6 kg of water. Therefore, the molality of the solution is 0.5 mol/kg.
The molality of the solution is 0.5 mol/kg. This is calculated by dividing the number of moles of solute (3 mol glucose) by the mass of solvent in kilograms (6 kg water).
4 mol over 0.800 kg
Glucose solution is a homogeneous mixture because it is composed of glucose dissolved in water, making it uniform throughout.
The molality of a solution is calculated by dividing the moles of solute by the mass of the solvent in kilograms. In this case, the molality would be 0.5 mol/kg since there are 3 moles of glucose in 6 kg of water.
yes
A 30% glucose solution is purely glucose and water, though it is actually impossible to keep other contaminants out of it. To create a 30% solution of glucose, you take a fixed volume of water and add 30% of that value of glucose to the water. The amount of glucose is in grammes. For example, 3g of glucose would be added to 10ml of water.
Dissolve 10 g pure glucose in 100 mL distilled water.
Glucose