1. How to prepare 15ml of 20% w/v NaCl solution? 2. How many...

1. How to prepare 15ml of 20% w/v NaCl solution? 2. How many grams of NaCl are required to prepare 45g of NaCl solution (35% w/w)? 3. Calculate the molarity of 0.289 moles of FeCl3 dissolved in 120 ml of solution. 4. How many moles of NaCl are present in 600 ml of a 1.55 M NaCl solution? 5. What is the molarity of 650 ml of solution containing 63 grams of NaCl? 6. How many grams of CaCO3 are needed to produce 500 ml of 1.66 M CaCO3 solution? How to prepare this solution? 7. What volume of a 0.88 M solution can be made using 130 grams of FeCl2? 8. Prepare 70ml of 1.5M NaCl solution from a 3M solution. 9. What is the volume of a 2 M H2SO4 solution prepared using 3 ml of 6 M H2SO4 solution? 10. What is the molarity of a 17 ml solution used to prepare 65 ml of 2.4 M solution?

Transcript

00:01 Hello student, here the question is given and the question is regarding the different solution we need to prepare and here the given value we already written.

00:09 So, the first one is the 15 ml of these 20 % weight by volume nacl solution we need to be prepared.

00:16 So, we have to calculate the amount of nacl required.

00:20 So, as already given here the 20 % weight by volume.

00:25 So, simply means these 20 gram in 100 ml of water.

00:32 So, this 20 gram present in 100 ml of solution of these nacl.

00:37 So, how much amount of these nacl present in the 15 ml.

00:42 So, it must be less than these 20 gram because 100 ml containing only 20 gram.

00:48 So, the 15 ml must be.

00:50 So, here we can say this 20 gram present in 100 ml.

00:57 So, by applying unitary methods we can say the 100 ml having these 20 gram.

01:03 So, the 1 ml is going to be this 20 gram divided by 100 ml.

01:10 So, this much amount of and similarly here we are taking this 15 ml.

01:16 So, this could be the 20 into 15 divided by 100.

01:20 So, by simply multiplying this we are getting the value of 3 gram of nacl.

01:27 So, here the answer should be this 3 gram of nacl is required and the second question is regarding this 45 gram of nacl is prepared and which is the given one is the 35 % weight by weight.

01:41 So, this 35 % weight by weight simply means 35 gram nacl in 100 ml 100 gram of solution.

01:53 So, similarly as we earlier read here it is 100 gram which is containing 35.

01:58 So, the one is going to be 35 by 100 and for the 45 gram we need this 35 divided by 100 into 45 and it's come around this 15 .75 gram.

02:14 So, the 15 .75 gram of these nacl is required in order to make this 45 gram.

02:24 So, this one is the answer for the second question and the similarly if we are calculating this molarity of this fecl3.

02:30 So, here the moles is given and the volume is already given and we know the molarity can be calculated by simply moles by volume and this volume should be in liter.

02:43 So, we need to convert this ml into liter by dividing the 1000.

02:49 So, here we are getting 0 .12 liter.

02:52 So, similarly 0 .28 putting the value and we are getting the molarity of this 2 .408 molar solution.

03:02 So, this is the answer and the next question is regarding the moles we have to calculate.

03:07 So, here also this applying the same formula.

03:11 So, this moles need to be calculated and this the volume in the liter here also this could be converted into the liter.

03:20 So, this is going to be 0 .6 liter and if we are calculating this moles this molarity divided into the volume...

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