# tactiq.io free youtube transcript
# CHEMISTRY 201: Solutions - Converting between Percent By Mass and Molarity
# https://www.youtube.com/watch/iYh2DEqrGU0

00:00:00.980 in this example problem I'll convert
00:00:03.889 between two units of concentration for
00:00:06.259 an aqueous solution percent by mass and
00:00:09.320 molarity our problem reads in HCl
00:00:12.980 solution is 26% HCl by mass and has a
00:00:16.460 density of 1.125 grams per milliliter
00:00:19.759 what is the molarity first we'll look at
00:00:23.210 the equation for molarity and molarity
00:00:25.970 equals moles of solute divided by liters
00:00:28.910 of solution next we'll look at the
00:00:32.360 equation for the units were given for
00:00:34.760 concentration percent by mass is equal
00:00:38.210 to the mass of the solute divided by the
00:00:40.609 mass of solution times 100% in this
00:00:44.870 problem I'm not given any information
00:00:46.960 regarding the amount of substance that I
00:00:50.090 have so I can make any one assumption
00:00:52.940 about the amount of substance I'm going
00:00:55.969 to make an assumption based on the
00:00:57.829 denominator of the equation for the
00:01:00.530 units that I have for concentration
00:01:02.800 because I'm starting with percent by
00:01:05.120 mass I'll make an assumption about the
00:01:07.640 mass of the solution to make the numbers
00:01:10.670 easy to work with
00:01:11.690 I'll assume I have 100 grams of solution
00:01:17.400 now that we've made an assumption about
00:01:19.650 the amount of solution that we have we
00:01:22.020 can calculate the moles of solute if we
00:01:25.410 rearrange our percent by mass equation
00:01:27.660 we'll get the mass of solute equals the
00:01:30.330 mass percent divided by 100 percent
00:01:33.210 times the mass of solution plugging
00:01:36.690 these numbers into our equation we'll
00:01:38.610 get 26 grams of HCL next we can convert
00:01:42.720 our grams of HCl into moles of HCl using
00:01:46.140 dimensional analysis our molar mass of
00:01:49.140 HCl is thirty six point four five 8
00:01:51.660 grams of HCL per mole and using
00:01:54.900 dimensional analysis we get zero point
00:01:57.090 seven one three moles of HCl
00:02:00.860 we now have calculated the numerator in
00:02:03.650 our molarity equation and now we need to
00:02:06.380 calculate liters of solution in the
00:02:08.479 denominator we can use dimensional
00:02:11.420 analysis and convert our 100 grams of
00:02:14.540 solution into liters of solution using
00:02:17.630 the given density of 1.125 grams per
00:02:21.140 milliliter when we put these numbers
00:02:23.540 into our calculator should get zero
00:02:25.790 point zero eight eight eight nine liters
00:02:28.400 of solution now we can plug our moles of
00:02:31.220 HCl and our liters of solution into our
00:02:34.070 molarity equation and we'll get eight
00:02:36.620 molar HCL

Quiz no. 3 ( 1 whole- intermediate only).

After watching the video,  you come with your own sample problems parallel to the one shown in the video. Just change the quantities 

Also,  answer question nos. 43 to 54 which can be found on the next screen shot.

Percent Solutions

(43) Calculate the mass percent of the following solutions:

(a) 25.0 g NaBr + 100.0 g H₂O

(b) 1.2g*K_{2}*S*O_{4} + 10g*H_{2}*O

45. How many grams of a solution that is 12.5% by mass AgNO, would contain 30.0 g of AgNO3?

Paired Exercises

48. Calculate the mass percent of the following solations

(a) 145.0 g NaOH in 1.5 kg H_{2}*O

44. Calculate the mass percent of the following solutions:

(a) 40.0 g Mg(NO 1 ) 2 + 500g*H_{2}*O (b) 17.5gN*O_{3} + 250g*H_{2}*O

46. How many grams of a solution that is 12.5% by mass AgNO, would contain 0.400 mol of Ag * N*O_{3}'

(b) 1.0 m solution of C_{6}*H_{12}*O_{6} in water

50. How much solute is present in 250. g of 15% K_{2}*Cr*C_{2} solution?

52. Calculate the mass/volume percent of a solution ade by dissolving 4.20 g of NaCl in H_{2}*O to make 125 ml. of solution.

54. What is the volume percent of 2.0 mL of hexane. C_{14r} dissolved in benzene, C_{6}*H_{6} to a voluine of 9.0 mi?