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The total volume of seawater is $1.5 \times 10^{21}$ L. Assume that seawater contains 3.1 percent sodium chloride by mass and that its density is $1.03 \mathrm{g} / \mathrm{mL} .$ Calculate the total mass of sodium chloride in kilograms and in tons $(1 \text { ton }=2000 \mathrm{lb} ; 1 \mathrm{lb}=453.6 \mathrm{g})$

First, let's figure out a few masses. We'll figure out the mass of the solution but taking the difference in mass of the evaporating dish before and after adding a solution. 12.5 grams. Then we can figure out the mass of any C L after evaporation by taking the mass of the n a C l that stuck after all the water evaporated minus just the massive evaporating dish would get 1.4 grams and then we have to figure out mass percent. We'll take the mass of any CL divided by the mass of the solution times 100. We get 11.6% in a C l. The density of the solution could be determined, which will need for part B by taking the mass of the solution, which we determined to be 12.5 grams, divided by the volume of the solution which was given to us at 10 milliliters and we get 1.205 grams for middle leader. So if we start with what we know, the grams of the sodium chloride divided by the grams of the solution, we need to convert the grams of sodium chloride into mold sodium chloride and then the mass of the solution into, um, leaders of solution. And we will get more clarity. So to convert the gram sodium chloride into mall sodium chloride will use it smaller Mass. Then to convert through the mass of the solution into mill. Leaders of the solution will use the density of the solution, which is 1.205 grams per mil leader. And then we don't want middle leaves in the denominator. We want leaders in the denominator. So will on the tail end here. Convert leaders are milliliters of solution into leader solution, and we'll get 2.40 Moeller sodium chloride, then last of all, we want to convert the similarity that we just calculated 2.40 into a new mill. Arat e. When we take 10 milliliters and we dilute it to 60 milliliters, then we can use the delusion equation and we'll solve for X and X in this case will be three 0.99 Moeller sodium chloride if we don't do any rounding it all until the very end

Yeah. We have a graduated cylinder that is weighed at 66 5 grounds. Mhm. And to the cylinder was added some water. Yeah. Has a mass of 58.2 g and some sodium chloride. Because a massive 5.279 grams. What is the total mass of the cylinder? And the solution? This is equal to 66.5 g plus 58.2 g. 5.279 g. We're gonna report our answer to one decimal place. It's yeah. One decimal place. We would get. 100 30.0 g. Would be the mask of the cylinder and solution. Right. Mhm.

For the total morality of all ions and seawater, we first need to convert every concentration value in parts per 1,000,000 into morality. Parts per 1,000,000 can also be expressed as milligrams per leader of a certain Saul you we can take advantage of this fact and you that convert every parts per 1,000,000 value into moles per kilogram of solvent. The first step is converting milligrams in two grams. The second step is converting grams to moles. The last step is converting leaders of solvent into kilograms of solvent. This equates to a morality value of 0.536 moles her killer Graham her chlorine. The same process will be done for sodium using sodium Smolar mass and the morality of sodium equates to 0.457 most per kilogram. I do then herself eight and then from magnesium and then for calcium, but none for potassium and then for bicarbonate. And lastly for a broom MyDD. Now, if we add all of these individual moralities up, the total morality for all ions in the seawater equates to 1.11 rolls her kilogram of water now in part B were asked to find the osmotic pressure at 300 Kelvin's of seawater, assuming Morality and Mueller tear equal. And if we substitute our morality, the gas constant and the temperature into the equation for osmotic pressure, it yields 27.3 atmospheres.

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