Can u answer this Question
To find the molarity, first calculate the number of moles of glucose in 40 g using its molar mass. Then, divide the moles of glucose by the volume of the solution in liters (1.5 L) to get the molarity.
To calculate the osmolarity of a solution containing 50mM of glucose, you'll need to consider the number of particles in solution. Glucose does not dissociate into multiple particles in solution, so its osmolarity is equivalent to its molarity. Therefore, the osmolarity of a 50mM glucose solution would be 50 mOsm/L.
To find the molarity of the sugar solution, we first need to convert the mass of glucose to moles. The molar mass of glucose (C6H12O6) is 180 g/mol. So, 80 g of glucose is equal to 80/180 = 0.444 moles. Next, calculate the molarity by dividing moles of glucose by liters of solution (0.75 L), Molarity = 0.444 moles / 0.75 L = 0.592 M.
The molarity of a solution with 3 mol of glucose in 6 kg of water cannot be determined without knowing the volume of the solution. Molarity is defined as the amount of solute (in mol) divided by the volume of the solution in liters. Without the volume, the molarity of the solution cannot be calculated.
2 moles in 2 liters means 1 mole in 1 liter - so it its molarity is 1.
To find the molarity, first calculate the number of moles of glucose in 40 g using its molar mass. Then, divide the moles of glucose by the volume of the solution in liters (1.5 L) to get the molarity.
To calculate the osmolarity of a solution containing 50mM of glucose, you'll need to consider the number of particles in solution. Glucose does not dissociate into multiple particles in solution, so its osmolarity is equivalent to its molarity. Therefore, the osmolarity of a 50mM glucose solution would be 50 mOsm/L.
To find the molarity of the sugar solution, we first need to convert the mass of glucose to moles. The molar mass of glucose (C6H12O6) is 180 g/mol. So, 80 g of glucose is equal to 80/180 = 0.444 moles. Next, calculate the molarity by dividing moles of glucose by liters of solution (0.75 L), Molarity = 0.444 moles / 0.75 L = 0.592 M.
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 ---------------------------
Find moles of glucose. 32.8 grams C6H12O6 (1 mole C6H12O6/180.156 grams) = 0.18206 moles C6H12O6 Molarity = moles of solute/Liters of solution M = 0.18206 moles C6H12O6/1.0 L = 0.18 M C6H12O6
To find the molarity (M) of the solution, use the formula: [ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} ] First, convert 183 mmol to moles: [ 183 , \text{mmol} = 0.183 , \text{mol} ] Next, convert 449 ml to liters: [ 449 , \text{ml} = 0.449 , \text{L} ] Now, calculate the molarity: [ \text{M} = \frac{0.183 , \text{mol}}{0.449 , \text{L}} \approx 0.408 , \text{M} ] Thus, the molarity of the solution is approximately 0.408 M.
The molarity of a solution with 3 mol of glucose in 6 kg of water cannot be determined without knowing the volume of the solution. Molarity is defined as the amount of solute (in mol) divided by the volume of the solution in liters. Without the volume, the molarity of the solution cannot be calculated.
2 moles in 2 liters means 1 mole in 1 liter - so it its molarity is 1.
4 mol over 0.800 kg
The first step is to calculate the number of moles of glucose by dividing the mass (15.8 g) by the molecular weight (180 g/mol). This gives 0.088 moles of glucose. Next, calculate the volume in liters by dividing 169 mL by 1000, which gives 0.169 L. Finally, divide the moles of solute by the volume of solution in liters to get the molarity: 0.088 moles / 0.169 L = 0.52 M.
Yes; the solution called "Ringer's Lactate" is a mixture containing water, essential salts and electrolytes, and glucose.
To calculate the osmolarity of a 5.5% glucose solution, first convert the percentage to grams per liter: 5.5% means 5.5 grams of glucose per 100 mL, which is 55 grams per liter. Glucose (C6H12O6) has a molar mass of approximately 180 g/mol, so 55 grams corresponds to about 0.306 moles per liter (55 g / 180 g/mol). Since glucose does not dissociate in solution, the osmolarity is equal to its molarity, which is approximately 0.306 osmoles per liter.