20%
98 mL
25 gallons
this question is not hard to answer, but it does require that one make some assumptions. The simplest answer assumes that the 20% salt solution refers to per cent by mass. Thus: a 20% salt solution is one which contains 20 mg salt for every 100 mg solution where the solution consists of a mixture of 20 mg salt plus 80 mg water. A 35% salt solution would contain 35 mg salt for every 65 mg water. Now, assuming that all the water in 18 mg of a 20% salt solution remains in the final solution we see that 18 mg salt solution x 20 mg salt/100 mg salt solution gives 3.6 mg of salt; thus, there are 18 mg total solution - 3.6 mg salt = 14.4 mg water. So the final salt solution must be one that contains 14.4 mg water and enough salt to make it 35% salt by mass. Mathematically, this is written as Z mg salt/(Z mg + 14.4 mg water) = 35/100 This gives Z = 0.35*(Z + 14.4) or Z = 0.35*Z + 0.35*14.4 which is same as Z = 0.35*Z + 5.04 and 0.65*Z = 5.04 so Z = 5.04/0.65 = 7.75 mg total salt needed. We started with 3.6 mg salt, so we must add 7.75 -3.6 = 4.15 mg salt Check: 7.75 mg salt/(7.75 mg salt + 14.4 mg water) = 0.35 or 35% There you go! --assuming that much salt dissolves that amount of water!
10
20%
98 mL
20% salt solution is the equivalent of adding 200gr salt in a 800 ml (1000ml -200ml) of water. you now have one liter of a 20% solution.
25 gallons
Pure water has a higher freezing point than 20% salt water.
Ice forms when the temperature reaches 32 degrees Fahrenheit. When salt is added the temperature drops so that a 10 per cent salt solution freezes at 20 degrees Fahrenheit and a 20 percent solution freezes at 2 degrees. Salt dissolves into the liquid water in the ice and lowers its freezing point
A pharmacist mixed a 20 percent solution with a 30 percent solution to obtain 100 liters of a 24 percent solution. How much of the 20 percent solution did the pharmacist use in the mixture (in liters).
this question is not hard to answer, but it does require that one make some assumptions. The simplest answer assumes that the 20% salt solution refers to per cent by mass. Thus: a 20% salt solution is one which contains 20 mg salt for every 100 mg solution where the solution consists of a mixture of 20 mg salt plus 80 mg water. A 35% salt solution would contain 35 mg salt for every 65 mg water. Now, assuming that all the water in 18 mg of a 20% salt solution remains in the final solution we see that 18 mg salt solution x 20 mg salt/100 mg salt solution gives 3.6 mg of salt; thus, there are 18 mg total solution - 3.6 mg salt = 14.4 mg water. So the final salt solution must be one that contains 14.4 mg water and enough salt to make it 35% salt by mass. Mathematically, this is written as Z mg salt/(Z mg + 14.4 mg water) = 35/100 This gives Z = 0.35*(Z + 14.4) or Z = 0.35*Z + 0.35*14.4 which is same as Z = 0.35*Z + 5.04 and 0.65*Z = 5.04 so Z = 5.04/0.65 = 7.75 mg total salt needed. We started with 3.6 mg salt, so we must add 7.75 -3.6 = 4.15 mg salt Check: 7.75 mg salt/(7.75 mg salt + 14.4 mg water) = 0.35 or 35% There you go! --assuming that much salt dissolves that amount of water!
Add 200 grams of the salt to 800 g of water, mix and dissolve, and you've got your desired 20% solution.
Mass solution=100g Mass solution=20g Mass Mass % = (mass Of solute/mass Of solution ) x 100. =(20/100)x100 =1/5 x 100. =100/5 =20%
- Sugar is not salt.- The unit of 20 is ?
10