Question

A teaspoon of table sugar contains about 0.01 mole sucrose what is the molarity of the superheroes in a cup of coffee is a teaspoon of sugar has been dissolved in a volume of 200 mL

Consider the sunion concentatton inNgat which is most appropriate a graph of the data above including all appropriate What is your independent variable which goes on the What is your dependent variable which goes on the y-axisi Procedure 3.3: Please complete the following solutions and dilution calculations (after lab is preferable) Qla A teaspoon teaspoon of sugar has been dissolved in a negar contains about 0.01 mol sucrose. What is the molarity of sucrose in a cup of coffee if a volume-o 200 mL 000 mol Oib: How much sugar (in moles) is contained in a modest sip (-10 mL) of the coffee mule Solu hion Q2x Calculate the M of 1000 mL common saline 9 g of Naci per l with molecular weight (MW) of NaCl = Q2b: You are working in an ER at a hospital and have run out of saline. You have sterile water and NaCI available to you. You need to make a stock (highly concentrated) solution of saline for patients coming into the ER. How many grams of NaCI are needed to make one liter of stock solution of saline a 25x the molarity above? Show your work below Procedure 3.4: Look up the pH of the following solutions using the following htaps/hwater.angs gowledulph.btm Battery Acid:E Acid Rain (range): Vinegar What is the most acidic substance above? Sea water Baking soda solution:0 Lemon juice Lye: lsye And the most basic H-1. hydrogen ion concentration (remember pH is a log scale) Students: please turn in I (one) graph attached to this worksheet, p. 5-6, when you enter the lab next week. Students should graph the volume of vinegar added (y-axis) vs. the solution molarity (x-axis).

Answer 1

Q1a.

Answer:

0.01 mol sucrose in 200 mL

Molarity?

Molarity= Moles/litres of solution

= 0.01/0.2L

=0.05moles/L

Q1b.

In 10mL 0f coffee

The concentration is 0.05moles/L

10mL ---> (10mL x 0.05moles)/1000mL =0.005 moles

Q2a.

Grams=Molarity/ (Volume x Molecular weight)

Molarity=(Grams x Molecular weight)/1L

= (9g)/(58.5x 1L)

=0.153M