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 prepare 0.02M NaOH from 1M NaOH solution, you will need to dilute the 1M solution. Use the formula: C1V1 = C2V2, where C1 is the concentration of the stock solution (1M), V1 is the volume of the stock solution you will use, C2 is the desired concentration (0.02M), and V2 is the final volume of the diluted solution. Calculate the volume of 1M NaOH solution (V1) needed to make the desired 0.02M concentration and dilute it with water to reach the desired volume (V2).
To make a 0.1M solution from a 1M HCL solution, you would dilute the 1M HCL with 10 parts of water (or whatever solvent you are using). For example, mix 1 mL of 1M HCL with 9 mL of water to obtain a 0.1M HCL solution.
Add 60g of Glacial Acetic Acid to a 1 liter volumetric flask. Make up to the mark with deionized water. The result is 1M acetic acid solution.
Glucose solution is a homogeneous mixture because it is composed of glucose dissolved in water, making it uniform throughout.
The ph. for this 1M Na2C4H2O4 solution can be found using the kA and the equation pH = pKa + log([base]/[acid]) This salt Na2C4H2O4 is going to increase the concentration of base in the solution.
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.
You could titrate equal volumes of 1M solution of NaOH and 1M solution of HCl to obtain 1M solution of NaCl.
A 1M solution of sodium carbonate means that it contains 1 mole of sodium carbonate dissolved in 1 liter of solvent (usually water). This concentration is used in chemistry to describe the amount of the solute (sodium carbonate) present in the solution.
To prepare 0.02M NaOH from 1M NaOH solution, you will need to dilute the 1M solution. Use the formula: C1V1 = C2V2, where C1 is the concentration of the stock solution (1M), V1 is the volume of the stock solution you will use, C2 is the desired concentration (0.02M), and V2 is the final volume of the diluted solution. Calculate the volume of 1M NaOH solution (V1) needed to make the desired 0.02M concentration and dilute it with water to reach the desired volume (V2).
To make a 0.1M solution from a 1M HCL solution, you would dilute the 1M HCL with 10 parts of water (or whatever solvent you are using). For example, mix 1 mL of 1M HCL with 9 mL of water to obtain a 0.1M HCL solution.
Add 60g of Glacial Acetic Acid to a 1 liter volumetric flask. Make up to the mark with deionized water. The result is 1M acetic acid solution.
Glucose solution is a homogeneous mixture because it is composed of glucose dissolved in water, making it uniform throughout.
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.
The ph. for this 1M Na2C4H2O4 solution can be found using the kA and the equation pH = pKa + log([base]/[acid]) This salt Na2C4H2O4 is going to increase the concentration of base in the solution.
To prepare 1M Tris-HCl from a 10mM solution, you would need to dilute the 10mM solution by a factor of 100. This means you would mix 1 part of the 10mM solution with 99 parts of water to achieve a final concentration of 1M Tris-HCl.
Mixing equal quantities of 1M HCl and 1M NaOH solutions will give a neutral solution because they will react to form water and a salt (NaCl).
No, glucose itself does not contain electrolytes. Electrolytes are typically ions such as sodium, potassium, chloride, and bicarbonate that can conduct electricity in solution. If electrolytes are needed, they would have to be added separately to a glucose solution.