percent concentration = (mass of solute/volume of solution) X 100
To solve for mass of solute,
mass of solute = (percent concentration X volume of solution)/100
So,
mass of solute = (10% X 100mL)/100 = 10g
3 percent alcohol
10g of K2CO3 in 100mL
Very simply 2g of ferric chloride and 100g (100ml) of water!
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.
it is O
3 percent alcohol
Weigh 10g of the given sample and dissolve it in 100ml of water, it forms a 10% solution of that sample.
10g of K2CO3 in 100mL
Well to make 30% stock solution of BSA from powder form you need to dissolve 30g of BSA into 100ml of solvent (water). Thus dissolved solution becomes 30% (w/v) stock solution.
0.06% W/V = 0.06g / 100ml = 60mg/100ml = 0.6mg/ml
Very simply 1g of ammonium thiocyanate and 100g (100ml) of water!
Very simply 2g of ferric chloride and 100g (100ml) of water!
add 25ml more of solution x * 20 = 100 * 25 x = 25
Unfortunately sugar is an imprecise term. It is better to specify glucose (usually means dextrose) or sucrose or fructose etc.A 5 percent solution of one of these sugars would contain 5 grams weight dissolved in 100mL of water (or could be another solvent).Read more: What_does_a_5_percent_sugar_solution_mean
Take 10 gm oF KI dissolve it in 100ml.
50 g
dissolve 10g of TCA in 100ml distilled water.