Depend on the kind of concentration it is expressed in. For a
1. 0.001 Molar solution, dissolve 0.001 mols of solute in enough solvent to obtain 1L of solution
2. 0.001 molal solution, dissolve 0.001 mols in 1kg of solution.
Molarity = moles of solute/Liters of solution ( 100 mL = 0.1 Liters)
Moles of solute (K2SO4) = Liters of solution * Molarity
Moles K2SO4 = 0.1 Liters * 0.1 M
= 0.01 moles K2SO4 (174.27 grams/1 mole K2SO4)
= 1.7 grams potassium sulfate
=======================Add that many grams potassium sulfate to your 100 mL.
Dissolve 0.745 grams of KCl in 100 ml of water to make 0.1M solution of KCl See the Related Questions link "How do you prepare a solution of a specific concentration?"
Assuming your compound has the same mass as water, take 0.01 g of compound in 100 ml solution, that will be 0.01 % solution of that compound.
So you have to have a milligram for every millilitre. But you have 100ml. Therefore you have to multiply 1mg by 100, to get 100mg. Weigh out 100mg or 0.1g and dissolve it to 100ml.
0.15g of leishman powder dissolved in 100ml of distilled water
6g Tris HCl + 100ml dH2O, pH 6.8
to prepare 100ml of 100mM Trissolution: Mol wt of Tris=121.14121.14g in 1000ml ----> 1M12.11g in 100ml -------->1M1M=1000mM121.1g---->1000mM12.11g ----------->100mM1.211g in 100ml and 100mM Tris
I don't think you can. The maximum solubility of glucose in water is 91% w/v. That would mean dissolving 91g of glucose in 100ml of water. for a 100% solution you would need to dissolve 100g in 100ml, and you cant do it under normal conditions.
To prepare a 3% acetic (ethanoic) acid solution, you must first standardise the ethanoic acid. This question assumes that has already been done. To make the concentration 3%, there must be 30g per 1000mL (30gL-1). Dissolve or mix in 30g of ethanoic acid per 1000ml (1L). Pipette this into aliquots of 100mL samples. You now have a 3% ethanoic/acetic acid solution.
Take 10 gm oF KI dissolve it in 100ml.
0.15g of leishman powder dissolved in 100ml of distilled water
6g Tris HCl + 100ml dH2O, pH 6.8
It depends on the final solution Volume you want to prepare. For 100ml of a 6M NaCL solution, you add 35.1g of NaCl to water until you reach 100ml. Dissolve and autoclave for 15 mins.
Put 0.9g of Nacl in a beaker and add distilled water to make up to 100ml. that is 0.9% Nacl (Normal saline Solution) From Tade Olubunmi Ademola
to prepare 100ml of 100mM Trissolution: Mol wt of Tris=121.14121.14g in 1000ml ----> 1M12.11g in 100ml -------->1M1M=1000mM121.1g---->1000mM12.11g ----------->100mM1.211g in 100ml and 100mM Tris
It means that 100mL of solution has 75g of solute dissolved in it.
I don't think you can. The maximum solubility of glucose in water is 91% w/v. That would mean dissolving 91g of glucose in 100ml of water. for a 100% solution you would need to dissolve 100g in 100ml, and you cant do it under normal conditions.
Solution A: Bismuth nitrate (0.17g) in AcOH (2mL) and H2O (8mL) Solution B: KI (4g) in AcOH (10mL) and H2O (20mL) Mix Solns. A and B and dilute to 100mL with H2O.
10g/200ml=5g/100ml = 5% solution. Now read your Dosage and Calculations book and prepare for the test because it is Monday. Good luck!
To prepare a 3% acetic (ethanoic) acid solution, you must first standardise the ethanoic acid. This question assumes that has already been done. To make the concentration 3%, there must be 30g per 1000mL (30gL-1). Dissolve or mix in 30g of ethanoic acid per 1000ml (1L). Pipette this into aliquots of 100mL samples. You now have a 3% ethanoic/acetic acid solution.
100ppm means, you need 100mg in 1oooml. Thus, for 100ml solution, you will need 10mg of methanol to make 100ppm solution.