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The molar mass of lactose is 342,3 g/mol; the solubility of lactose is 216 g/L at
20 0C. Consequently you cannot prepare a molar solution of lactose.
Molarity = moles of solute / L of solutionOsmolarity = osmoles of solute / L of solution1 Osmole = 1 mole of osmotically-active solute. (NH4)3PO4 dissociates into 3 NH4+ ions and 1 PO4- ion (total 4 solute particles), so 4 osmoles of (NH4)3PO4 result from 1 mole of (NH4)3PO4. Therefore, because 0.4 moles (NH4)3PO4 per Liter solution are in an 0.4 M solution, 1.6 osmoles of (NH4)3PO4 per Liter solution are present. Thus...Osmolarity = osmoles of solute / L of solution = 1.6 osmoles of (NH4)3PO4 / L of solution = 1.6 OsM
A mole typically grows to be about 6 inches in length, including the tail.
A mole on the skin is made up of clusters of pigment-producing cells called melanocytes. These cells give the mole its color and can sometimes grow in a raised or flat shape on the skin.
Having hair on a mole can make it difficult to monitor changes in the mole's appearance, which could be a risk factor for skin cancer. It is recommended to consult a dermatologist to determine if the mole should be removed for health reasons.
There are 6.022 x 10^23 molecules in 1 mole of glucose. This number is known as Avogadro's number and represents the number of units (atoms, molecules, etc.) in one mole of a substance.
342 grams of Lactose in 1 litre water or 34.2 grams in 100 mls.
To prepare a 1 mole solution of dimethoxyhydroxyacetophenone, you would dissolve 166.21 grams of the compound in enough solvent to make a total volume of 1 liter. Calculate the required weight based on the molar mass of dimethoxyhydroxyacetophenone (C10H12O4).
mole of lactose x 6.023 x 10(>23) molecules (avagadro number)
Dissolve 1.0 mole gas (17 gram) in 1.0 Liter water
No, the mole of solution is not equal to the mole of solute plus the mole of solvent. The mole of solution refers to the total amount of moles in a given volume of solution, which includes both the solute and the solvent.
To prepare a 1N NaOH solution, you would need to dissolve 40 grams of NaOH in water to make 1 liter of solution. This amount is used because 1N solution means 1 mole of NaOH per liter of solution, and the molar mass of NaOH is 40 g/mol, so 40 grams of NaOH is needed to have 1 mole in 1 liter of solution.
You could titrate equal volumes of 1M solution of NaOH and 1M solution of HCl to obtain 1M solution of NaCl.
Ammonium molybdate is (NH4)2MoO4. Its molar mass is 148g/mol. A 0.5M solution will have half a mole per liter of water. Half a mole of ammonium molybdate is 74g, so you would measure out 74g of it, and dissolve it in a liter of water.
To prepare a 10ml solution containing 1 mole of BSA (Molecular weight 66000 g/mol), you would need 66g of BSA. To prepare a 100ml solution containing 1 mole of BSA, you would need 660g of BSA. Remember to adjust the volume accordingly after dissolving the BSA to ensure accurate concentration.
To prepare a 0.01M KCl (potassium chloride) solution in 1 liter, you would need to dissolve 0.74 grams of KCl in enough water to make 1 liter of solution. This can be calculated using the formula: moles = Molarity x Volume (in liters) x Molecular weight of KCl.
One mole solution of sodium chloride makes 1000 millimole. So 0.1 mole solution of sodium chloride will have 100 millimole in the solution.
To prepare a 1 M solution of perchloric acid (HClO4), you would need to dissolve 1 mole of HClO4 in enough water to make a final volume of 1 liter. This can be calculated by using the formula: moles = Molarity (M) x Volume (L). Please handle perchloric acid with caution and use appropriate safety measures while preparing the solution.